9/20/20151 numerical solution of partial differential equations using compactly supported radial...
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04/19/23 1
Numerical Solution of Partial Differential Equations Using Compactly Supported Radial Basis Functions
C.S. Chen
Department of Mathematics
University of Southern Mississippi
04/19/23 2
PURPOSE OF THE LECTUREPURPOSE OF THE LECTUREPURPOSE OF THE LECTUREPURPOSE OF THE LECTURE
TO SHOW HOW COMPACTLY SUPPORTED RADIAL
BASIS FUNCTIONS (CS-RBFs) CAN BE USED TO
EXTEND THE APPLICABILITY OF BOUNDARY
METHODS TO PROVIDE 'MESH FREE' METHODS
FOR THE NUMERICAL SOLUTION OF PARTIAL
DIFFERENTIAL EQUATIONS.
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04/19/23 4
MESH METHODMESHLESS METHOD
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The Method of Particular Solutions
Consider the Poisson’s equation
,),( xxfLu,),( xxgu
Where ,3,2, dRd is a bounded open nonempty domain
with sufficiently regular boundary .Let puuv where pu satisfying )(xfLu p but does not necessary satisfy the boundary condition in (2).
(1)(2)
v satisfies , ,0 xLv
. ),()( xxx pugu
(3)
(4)(5)
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• Domain integral Domain integral
• Atkinson’s method (C.S. Chen, M.A. Golberg & Y.C. Hon, The MFS and quasi-Monte Carlo method for diffusion equations, Int. J. Num. Meth. Eng. 43,1421-1435, 1998)
(Requires no meshing if = circle or sphere)
• Radial Basis Function Approximation of
• Others
)();,()( dvQfQPGPu p
for solution lfundamenta a is );,( QPG L
ˆ,)();,()( dvPfQPGPu p
f
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Assume that )(ˆ)( xx ff and that we can obtain an analytical solution up
to
).(ˆˆ xfu p Then
.ˆ pp uu
To approximate f by f we usually require fitting the given
data set xi
N
1 of pairwise distinct centres with the imposed
conditions
.1 ),(ˆ)( Niff xx
The Method of Particular Solutions
(6)
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The linear system
( ) , ,f a i Ni ii
N
i jx x x
1
1
is well-posed if the interpolation matrix is non-singular
A i ji N
x x1
Once f in (6) has been established,
u ap i ii
N
1
(7)
(8)
where i iand
i i i i x x x x, .
(9)
04/19/23 9
Compactly Supported RBFsLet be a continous function with (0) 0. If let : ,R R i
x
i i ix x x ,
where is the Euclidean norm. Then is called the RBF. i
References• Z. Wu, Multivariate compactly supported positive definite radial functions, Adv. Comput. Math., 4, pp. 283-292, 1995.• R. Schaback, Creating surfaces from scattered data using radial basis functions, in Mathematical Methods for Curves and Surface, eds. M. Dahlen, T. Lyche and L. Schumaker, Vanderbilt Univ. Press, Nashville, pp. 477-496, 1995• W. Wendland, Piecewise polynomial, positive definite and compactly supported RBFs of minumal degree, Adv. Comput. Math., 4, pp 389- 396, 1995.
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• C.S. Chen, C.A. Brebbia and H. Power, Dual receiprocity method using compactly supported radial basis functions, Communications in Numerical Methods in Engineering, 15, 1999, 137-150.
• C.S. Chen, M. Marcozzi and S. Choi, The method of fundamental solutions and compactly supported radial basis functions - a meshless approach to 3D problems, Boundary Element Methods XXI, eds. C.A. Brebbia, H. Power, WIT Press, Boston, Southampton, pp. 561-570, 1999.
• C.S. Chen, M.A. Golberg, R.S. Schaback, Recent developments of the dual reciprocity method using compactly supported radial basis functions, in: Transformation of Domain Effects to the Boundary, ed. Y.F. Rashed, WIT PRESS, pp183-225, 2003.
• M.A. Golberg, C.S. Chen, M. Ganesh, Particular solutions of the 3D modified Helmholtz equation using compactly supported radial basis functions, Engineering Analysis with Boundary Elements, 24, pp. 539-547, 2000.
References
04/19/23 11
Wendland’s CS-RBFs
( )( ) , ,
, .1
1 0 1
0 1
rr r
rn
nDefine
For d=1,
( )
( ) ( )
( ) (8 )
1
1 3 1
1 5 1
0
3 2
5 2 4
r C
r r C
r r r C
For
For d=2, 3,
( )
( ) ( )
( ) ( )
( ) ( )
1
1 4 1
1 35 18 3
1 32 25 8 1
2 0
4 2
6 2 4
8 3 2 6
r C
r r C
r r r C
r r r r C
(10)
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Globally Supported RBFs
φ=1+r r c2 2
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Compactly Supported RBFs
( )1 2r ( ) ( )1 4 14r r
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Compact support cut-off parameter (scaling factor)
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Analytic Particular Solutions L= in 2D
r
r rr
r
1 4 1
0
4
, ,
, .
1, in 2D,
d dr
r dr dr
4
7 6 5 4 2
5 4 3 2
= 1 4 1
4 5 4 5, .
7 2 2 2
d r r rr r dr
dr
r r r r rA r
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6 5 4 3
15 4 3 2
7 6 5 4 2
5 4 3 2
4 5 4 5=
7 2 2 2
4 5 4 5, .
49 12 5 8 4
r r r r r rC dr
r r r r r AB r
r
where C and D are to be determined by matching and ' at .r
Choose A= B = 0, we have
7 6 5 4 2
5 4 3 2
4 5 4 5, ,
= (*)49 12 5 8 4ln( ) , ,
r r r r rrr
C r D r
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6 5 4 3
5 4 3 2
4 5 4 5= , ,
7 2 2 2
,
d r r r r r rr
drC
rr
Note that
6 5 4 3
5 4 3 2
2
4 5 4 5For ,
7 2 2 2
14 14
Cr
CC
From (*), we have 2 2529
ln5880 14
D
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7 6 5 4 2
5 4 3 2
2 2
4 5 4 5, ,
49 12 5 8 4=
529ln , ,
5880 14
r r r r rr
r
rr
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Example in 2D
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25( , ) sin( ) cos , ( , ) ,
4 2
( , ) sin( ) cos , ( , ) .2
yu x y x x y
yu x y x x y
Example 2D
-1 -0.5 0 0.5 1
-1
-0.5
0
0.5
1
-1 -0.5 0 0.5 1
-1
-0.5
0
0.5
1
10,000 uniform grid points
04/19/23 23Sparse matrix for uniform grid points.
04/19/23 24
CS-RBF used:
4
1 4 1r r
MFS: Fictitious circle with r = 4.
Interpolation points: 100x100 uniform grid points
α RMSE NZ C PU (sec)
0.10 3.76E-4 561,013 1.65
0.20 1.55E-5 2,171,057 3.36
0.25 8.43E-6 3,376,025 5.32
04/19/23 25
Analytic Particular Solutions L= in 3D
r
r rr
r
1 4 1
0
4
, ,
, .
Recall
Since
12
2
r
d
drr
d
dr, in 3D,
we have
(11)
r
r r r r rr
rr
2 4
2
5
3
5
4
7
5
2 36 2
2
3
5
16 14
14 42
,
, .
04/19/23 26
Numerical Example in 3D
Consider the following Poisson’s problem
u x y z x y z x y z
u x y z x y z x y z
( , , ) cos( ) cos( ) cos( ), ( , , ) ,
( , , ) cos( ) cos( ) cos( ), ( , , ) .
3
-1
0
1-0.5-0.2500.250.5-0.5
-0.25
0
0.25
0.5
-1
0
1
-0.5
-0.25
0
0.25
0.5
Physical Domain
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-1.0 -0.5 0.0 0.5 1.0 1.50
1
2
3
4
5
6
7
8
= 1.4
= 1.0
= 0.7
Rel
ativ
e E
rror
s (
%)
X Axis
The effect of various scaling factor α
We choose 1 4 14r r( ) to approximate the forcing term.
To evaluate particular solutions, we choose 300 quasi-random pointsin a box [-1.5,1.5]x[-0.5,0.5]x[-0.5,0.5]. The numerical results are compute along the x-axis with y=z=0.
04/19/23 28
Modified Helmholtz Equation in 3D
2 22
1 ( ), 0 ,1
0, .
nr
p r rd dr
r dr drr
Let j ( )( )
, , ( ) lim , , , .rw r
rr
d
dr
w
rj
r
j
j
0 0 0 1 2
0
1 12
22
2r
d
drr
d
dr r
d w
dr
(16)
(15)
(17)
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Hence, (15) is equivalent to
d w
drw r
rp
rr
r
n2
22 1 0
0
, ,
, .
The general solution of (18) is of the form
w rAe Be q r r
Ce De r
r r
r r( )( ), ,
, .
0
q(r) can be obtained by the method of undetermined coefficients,or by symbolic ODE solver (MATHEMATICA or MAPLE).A,B,C and D in (19) are to be chosen so that in (16) is twice differentiable at r = 0 and
(18)
(19)
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Theorem 1. Let w be a solution of (18) with w(0)=0. Then defined by (16) is twice continuously differentiable at 0 with (0)=w'(0), '(0)=0, ''(0)=[s2 w'(0)+p(0)]/3.Furthermore, (r) satisfies (15) as limr→0+
A B q
Ae Be q Ce De
A e B e q C e D e
( )
( )
' ( )
0 0
Choose D = 0. Consequently, we get
A B q
Be q q
C B e q e q
[ ( )]( ' ( ) ( )]
( ) ( ) ( )
0
21 02
(20)
(21)
04/19/23 31
Hence, the particular solution Φ is given by
( )
( ( ) ' ( ), ,( )
, ,
, .
r
s B q q rAe Be q r
rr
Ce
rr
sr sr
sr
2 0 0 0
0
Notice that for r > α and large wave number λ
Ce
r
e q q
r
q e q e
rr
r r r r
( ) ( )' ( ) ( ) ( )
.2
0
0 for
(22)
04/19/23 32
d w
drw r
rr
2
22
2
1 0
,
Example
The general solution of
is given by
w rr r r r
Ae Ber r( )
4 6 22 2 2 2 3 2
4 2
q rr r r r
( ) 4 6 22 2 2 2 3 2
4 2
Hence,
A,B,C can be obtain from (21).
04/19/23 33
For
( ) ,rr r
1 4 1
4
q r r
r r
r r r
( )
480 2880 60 1800 1
240 1400 10 300
20 120 15 4
6 3 8 5 4 2 6 4 2
24 3 6 5
32 2 4 4
42 3 4 5 2 4
52 5
6
04/19/23 34
Numerical Results
400I u f
u g
, ,
, .
in
on
( , , ) : ( , , )x y z R H x y z3 1 with
H x y z x x y z( , , ) min ,
3
4
3
4
2 2
2 2
Consider (23)(24)
04/19/23 35
We choose f and g such that the u(x,y,z)=ex+y+z /400 is theexact solution of (23)-(24).To evaluate particular solutions, we choose N=400 quasi-random points . We choose the CS-RBFs
( )rr
12
04/19/23 36
Error estimates for various compact support cut-off parameter
||u-uN || CPU-time (sec.)
0.2 2.2864E-02 01.40
0.4 1.6194E-02 04.43
0.6 9.7795E-03 07.48
0.8 6.4514E-03 10.63
1.0 4.5061E-03 13.71
1.2 3.8208E-03 16.98
1.4 3.3006E-03 20.53
1.6 2.9089E-03 24.08
1.8 2.6375E-03 27.60
04/19/23 37
Multilevel Interpolation
Given a set of X scattered points, we decompose X into a nestedsequence
X X X XM1 2 M subsets X x x X k Mk
kNk
k 1 1( ) ( ), , , .
Algorithm:
S f
S f S
S f S
X X
X X
M X kk
M
XM M
1
2 1
1
1
1 1
2 2
| | ,
| ( )| ,
| ( )| .
S x c x xk jk
k jk
j
N k
( ) ( ) ( )
1
S S S fM X X1 2 | |
04/19/23 38
The basic idea is to set α1 relatively large with few interpolationpoints, and to let the αk decrease as k increases with more points.
Reference:M.S. Floater, A. Iske, Multistep scattered data interpolation using compactlysupported radial basis functions, J. Comp. and Appl. Math., 73, 65-78, 1996.
30 interpolation points 100 interpolation points
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300 interpolation points
04/19/23 40
Numerical Results
u e x y D
u e e y x y D
D
x y
x y x
2 , ( , ) ,
cos , ( , ) ,
where is the Oval of Cassinii.
10.50-0.5-1
0.4
0.2
0-0.2
-0.4
04/19/23 41
x y Numer. Exact Abs. Error
0.00 0.00 1.99994 2.00000 0.000061
-0.63 0.00 3.74097 3.74098 0.000012
1.34 0.00 7.65247 7.65261 0.000028
0.17 0.17 2.17206 2.35207 0.000082
0.00 0.20 1.79712 1.79707 0.000052
-0.37 0.37 1.11578 1.11594 0.000156
-1.34 0.00 0.5228 0.52270 0.000130
α/1.5, 1.0, 0.6, 0.2/ n/30, 60, 200, 400/
04/19/23 42
For time-dependent problems, we consider two approaches to convert problems to Helmholtz
equation • LAPLACE TRANSFORM
• FINITE DIFFERENCES IN TIME
ALSO POSSIBLE
• OPERATOR SPLITTING (RAMACHANDRAN AND BALANKRISHMAN)
04/19/23 43
DPPhPu ),0,()0,(
(HEAT EQUATION)
3,2,),,(),(),( dRDPtPftPutPu dt
0,),,(),( tDPtPgtPu
Consider the BVP
where D is a bounded domain in 2D and 3D.
Let
0
),(),(ˆ dttPuesPu st
(28)Then
(25)(26)(27)
),(ˆ sPu satisfies
DPsPgsPu
sPMPhsPfsPussPu
),,(ˆ),(ˆ
),()(),(ˆ),(ˆ),(ˆ (29)(30)
04/19/23 44
(WAVE EQUATION)
)()0,(),()0,(
0,),,(),(
0,),,(),(
00 PvPuPuPu
tDPtPgtPu
tDPtPutPu
t
tt
Consider BVP
Then ),(ˆ sPu satisfies
DPsPgsPu
PvPsusPussPu
),,(ˆ),(ˆ
)()(),(ˆ),(ˆ 002
Similar approach can be applied to hyperbolic-heat equation and heat equations with memory
(31)
(32)(33)
(34)(35)
04/19/23 45
(HEAT EQUATION)1For .0,Let nnn tttnnt
),()1(),(),(
),()1(),(),(
1
1
nn
nn
tPutPutPu
tPutPutPu
and/)],(),([),( 1 nnt tPutPutPu
satisfies ),()(Then nn tPuPv )(//)1(//11 Pmfvvvv nnnnnn
Method) s(Rothe' 1For nnnn fvvv //11
Nicholson)-(Crank 1/2For )(2/2/2 11 Pmfvvvv nnnnnn
(36)(37)
(38)
(39)
(40)
(41)
04/19/23 46
(J. Su & B. Tabarrok, A time-marching integral equation method for unsteady
state problems, Comp. Meths. Appl. Mech. Eng. 142, 203-214, 1997) (WAVE EQUATION)
211 /)2( nnntt uuuu
satisfies Then nn uv
12
12
11 )1(/)2(/ nnnnn vvvvv
CONVECTION-DIFFUSION EQUATION
2/)(
field) velocity (
11t
nn
t
uuu
VuuVu
nnnnn vVvvvv
2/)()1( 1111
Similar approach works for non-linear equation
),,(),(),( uuPftPutPu t
(41)
(42)
(43)(44)
(45)
(46)